Time-Optimal Control & Switching Surfaces

Minimum-time control is the cleanest place to see how the maximum principle becomes feedback geometry. With no running penalty beyond elapsed time, the optimizer uses all available authority except at isolated switches. The hard part is not guessing “full throttle”; it is finding when to reverse it so the state reaches the target with the required terminal velocity.

Double integrator: problem and PMP #

Normalize the acceleration bound to one:

$$\dot x_1=x_2,\qquad \dot x_2=u,\qquad |u|\le1,$$

with initial state $(x_{10},x_{20})$, target $(0,0)$, free terminal time, and

$$J=t_f-t_0=\int_{t_0}^{t_f}1\,dt.$$

Using $H=p_1x_2+p_2u-1$, the costate equations are

$$\dot p_1=0,qquad \dot p_2=-p_1.$$

Therefore

$$p_1=c_1,qquad p_2(t)=c_2-c_1t.$$

Hamiltonian maximization gives

$$u^*(t)=\mathop{\mathrm{sgn}}p_2(t)$$

away from a zero of $p_2$. A nonzero affine function has at most one zero. If it vanished identically, then $p_1=p_2=0$, violating nontriviality. Hence every normal time-optimal trajectory has at most one switch.

Constant-control phase curves #

On an interval with constant $u\in\{-1,+1\}$,

$$\frac{dx_2}{dx_1}=\frac{u}{x_2}.$$

Multiplying by $x_2$ and integrating gives

$$\frac12x_2^2=ux_1+C,$$

or

$$\boxed{x_2^2-2ux_1=C.}$$

The final arc must pass through the origin, so $C=0$. If $x_2>0$, braking requires $u=-1$ and the final arc is $x_1=-x_2^2/2$. If $x_2<0$, braking requires $u=+1$ and the final arc is $x_1=x_2^2/2$. Both branches are

$$\boxed{s(x)=x_1+\frac12x_2|x_2|=0.}$$

This is the switching curve.

Exact state-feedback law #

Off the switching curve, define

$$s(x)=x_1+\frac12x_2|x_2|.$$

The time-optimal feedback is

$$\boxed{u^*(x)=-\mathop{\mathrm{sgn}}s(x),\qquad s(x)\ne0.}$$

On $s(x)=0$ and away from the origin, take the final braking action

$$\boxed{u^*(x)=-\mathop{\mathrm{sgn}}x_2.}$$

At the origin the control can be set to zero. The on-curve clause matters because a generic implementation of $\mathop{\mathrm{sgn}}(0)$ does not encode which parabolic branch leads to the target.

Switching time and minimum travel time #

Let

$$\sigma=\mathop{\mathrm{sgn}}s(x_0)$$

for an initial state off the switching curve. The first control is $u_1=-\sigma$. Let its duration be $\tau$. At the switch,

$$x_{2s}=x_{20}-\sigma\tau,$$

$$x_{1s}=x_{10}+x_{20}\tau-\frac12\sigma\tau^2.$$

The second control is $u_2=\sigma$, and the switch point must lie on its final arc:

$$x_{1s}=\frac12\sigma x_{2s}^2.$$

Substitution produces

$$\tau^2-2\sigma x_{20}\tau +\frac12x_{20}^2-\sigma x_{10}=0.$$

The physically valid root is

$$\boxed{\tau=\sigma x_{20} +\sqrt{\frac12x_{20}^2+\sigma x_{10}}.}$$

At that instant,

$$x_{2s}=-\sigma \sqrt{\frac12x_{20}^2+\sigma x_{10}}.$$

The final braking duration is $|x_{2s}|$, hence

$$\boxed{T^*(x_0)=\sigma x_{20} +2\sqrt{\frac12x_{20}^2+\sigma x_{10}}, \qquad s(x_0)\ne0.}$$

If $s(x_0)=0$, the state is already on a final arc and

$$\boxed{T^*(x_0)=|x_{20}|.}$$

For an acceleration limit $|u|\le a_{\max}$, replace the switching curve by

$$x_1+\frac{x_2|x_2|}{2a_{\max}}=0$$

and scale each time formula consistently by $a_{\max}$.

Worked state #

Take $x_0=(10,0)$ and $|u|\le1$. Then $s=10$, so $\sigma=1$ and

$$\tau=\sqrt{10},\qquad x_s=(5,-\sqrt{10}).$$

The controller applies $u=-1$ for $\sqrt{10}$ seconds, then $u=+1$ for another $\sqrt{10}$ seconds. Thus

$$T^*=2\sqrt{10}.$$

Accelerating toward the origin until $x_1=0$ would be too late: the state would cross the origin with nonzero velocity. The switching parabola encodes the distance needed to cancel that velocity exactly.

Harmonic oscillator: circular arcs and repeated switches #

Consider

$$\dot x_1=x_2, \qquad \dot x_2=-\omega^2x_1+u, \qquad |u|\le u_{\max}.$$

For a constant $u$, shift the equilibrium to $x_1=u/\omega^2$. Direct differentiation shows that

$$E_u=x_2^2+\omega^2 \left(x_1-\frac{u}{\omega^2}\right)^2$$

is constant. Each saturated arc is therefore an ellipse in $(x_1,x_2)$, or a circle in the scaled plane $(\omega x_1,x_2)$, centered at $(u/\omega,0)$.

The costate satisfies

$$\dot p_1=\omega^2p_2, \qquad \dot p_2=-p_1,$$

and hence

$$\ddot p_2+\omega^2p_2=0.$$

Thus

$$p_2(t)=A\cos\omega t+B\sin\omega t,$$

and $u^*=u_{\max}\mathop{\mathrm{sgn}}p_2$. Consecutive simple zeros of $p_2$ are separated by $\pi/\omega$. Unlike the double integrator, the switching function can cross zero repeatedly, so the global switching set is a piecewise chain of alternating constant-control arcs.

Tracking is a coordinate translation, when the reference is an equilibrium #

Suppose the desired constant state $x_r$ is maintained by $u_r$:

$$0=Ax_r+Bu_r.$$

With $e=x-x_r$ and $v=u-u_r$,

$$\dot e=Ae+Bv.$$

The regulator switching geometry applies in $e$ coordinates, while the physical control is $u=v+u_r$. The admissible set also shifts: if $u\in[u_{\min},u_{\max}]$, then

$$v\in[u_{\min}-u_r,\,u_{\max}-u_r].$$

If that interval is asymmetric, the two constant-control arcs have different curvature and duration. Merely drawing the origin-centered symmetric curve around the reference would be wrong.

Implementation limits #

The ideal feedback is discontinuous. Real actuators introduce sampling, delay, rate limits, deadband, and unmodeled modes. A naive sampled sign law can chatter around the curve. A boundary layer or hysteresis may reduce switching, but it changes the exact minimum-time problem and creates a nonzero terminal region. Its width must be chosen from actuator and estimation limits, then the achieved time and terminal error must be verified on the real constrained model.